Dihedral and cyclic extensions with large class numbers Tome 24, no3 (2012), p. 583-603.
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de Bordeaux 24 (2012), 583-603
Dihedral and cyclic extensions with large class
numbers
par Peter J. CHO et Henry H. KIM
Résumé. Cet article est la suite de [2]. Nous construisons incon-ditionnellement plusieurs familles de corps de nombres ayant un grand nombre de classes. Ce sont des corps de nombres dont la clôture galoisienne a pour groupe de Galois les groupes dièdraux
Dn, n = 3, 4, 5, et les groupes cycliques Cn, n = 4, 5, 6. Nous
construisons d’abord des familles de corps de nombres à petits régulateurs et, en utilisant la conjecture d’Artin forte et en ap-pliquant une variante du résultat de densité nulle de Kowalski et Michel, nous choisissons des sous-familles telles que les fonctions L correspondantes soient sans zéro près de 1. Pour ces sous-familles, la fonction L prend une valeur extrémale en s = 1 et, par la for-mule du nombre de classes, nous obtenons un grand nombre de classes.
Abstract. This paper is a continuation of [2]. We construct unconditionally several families of number fields with large class numbers. They are number fields whose Galois closures have as the Galois groups, dihedral groups Dn, n = 3, 4, 5, and cyclic groups Cn, n = 4, 5, 6. We first construct families of number fields with small regulators, and by using the strong Artin conjecture and ap-plying some modification of zero density result of Kowalski-Michel, we choose subfamilies such that the corresponding L-functions are zero free close to 1. For these subfamilies, the L-functions have the extremal value at s = 1, and by the class number formula, we ob-tain large class numbers.
1. Introduction
This paper is a continuation of [2]. Let K(n, G, r1, r2) be the set of
num-ber fields of degree n with signature (r1, r2) whose normal closures have G
as their Galois group. In [2], we constructed families of number fields with the largest possible class numbers belonging to K(5, S5, 1, 2), K(4, S4, 2, 1),
Manuscrit reçu le 20 juillet 2011.
K(4, S4, 0, 2) and K(4, A4, 0, 2). In this paper, we construct families of
num-ber fields with large class numnum-bers whose Galois closures have as Galois groups, dihedral groups Dn, n = 3, 4, 5, and cyclic groups Cn, n = 4, 5, 6.
The idea is the same as in [2]. Namely, we use the class number formula; the class number hKt for Kt∈ K(n, G, r1, r2) is given by
hKt = wKt|dKt| 1 2 2r1(2π)r2RK t L(1, ρt),
where wKt is the number of roots of unity in Kt, dKt is the discriminant of Ktand RKt is its regulator and L(s, ρt) =
ζKt(s)
ζ(s) is the Artin L-function.
We first construct a family of number fields Kt with small regulators, and then find a subfamily for which L(1, ρt) (log log |dKt|)
n−1. However, in our cases, ρt is no longer irreducible, and hence it is no longer attached to a cuspidal automorphic representation. We need to modify the result of [14] to isobaric automorphic representations. See section 3.
In section 4, we use the quintic polynomial considered by Schöpp [22] and Lavallee-Spearman-Williams-Yang [15] who showed that they give rise to number fields whose Galois closures has D5 as the Galois group, and computed fundamental units when they are of signature (1,2). We prove that they give rise to regular Galois extensions over Q(t), and we compute regulators and show that they have the largest class numbers.
In section 5, we use the family of quartic polynomials considered by Nakamula [19] who showed that they give rise to number fields whose Galois closures have D4 as the Galois group, and computed regulators. We prove that they give rise to regular Galois extensions over Q(t), and show that they have large class numbers. We conjecture that they have the largest possible class numbers.
In section 6, we prove that the cubic extensions considered by Ishida [11] have the largest possible class numbers. This family is different from the one considered by Daileda [5].
In section 7, we consider totally real cyclic extensions. They are some-times called simplest fields. We prove a very general result that a cyclic extension of prime degree generated by a unit always has the smallest reg-ulator. We use the result in [7] that if the cyclic extension is given by a polynomial f (x, t) of degree p with the constant term ±1, then any p − 1 roots are multiplicatively independent. However, it is no longer true for composite degree. Cyclic quintic fields were studied by Lehmer [17], Jean-nin [12], and Schoof-Washington [21]. We compute the regulators and prove that they have the largest possible class numbers. Cyclic quartic fields were studied by Lazarus [16], and cyclic sextic fields by Gras [9]. They deter-mined fundamental units. We compute the regulators and prove that they have large class numbers.
When n = 4, 6, due to the existence of subfields, our regulator bounds are greater than Silverman’s bounds [25]. However, we conjecture that in the case of regular Galois extensions, our bounds are sharp. (see Remark 7.3.)
2. Representations of dihedral groups
Let’s review irreducible representations of Dn: If n is odd, Dn=< a, x :
an = x2 = e, xax = a−1 >. Let H = {1, x}. Then irreducible
represen-tations of Dn are: 2 one-dimensional representations 1, χ, and n−12 two-dimensional representations ρ1, ..., ρn−1
2
, where χ is the character of H. We have IndGH1 = 1 + ρ1+ · · · + ρn−1
2
.
If n is even, Dn =< a, x : an = x2 = e, xax = a−1 >. Let H 1 =
{1, x}, H2 = {1, an2}, H3 = {1, a
n
2x} be three order 2 subgroups. Then
irreducible representations of Dn are: 4 one-dimensional representations 1, χ1, χ2, χ3, and n−22 two-dimensional representations ρ1, ..., ρn−2
2
, where χi is the character of Hi. We have, for each i, IndGHi1 = 1+χi+ρ1+· · ·+ρn−22 .
Let K/Q be a degree n extension and ˆK/Q be the Galois closure such
that Gal( ˆK/Q) ' Dn. Then if n is odd,
ζK(s) ζ(s) = L(s, ρ1) · · · L(s, ρn−12 ). If n is even, ζK(s) ζ(s) = L(s, χ)L(s, ρ1) · · · L(s, ρn−22 ),
where H = Gal( ˆK/K) is one of the order 2 subgroups of Dn, and χ is the non-trivial character of H.
3. Approximation of L(1, ρ) and zero-free region
We use the following result of Daileda [5] to obtain a bound for L(1, ρ): Let ρ be an l-dimensional complex representation of a Galois group. We assume L(s, ρ) is an entire Artin L-function and N is its conductor. Also
L(s, ρ) has a Dirichlet series L(s, ρ) =
∞
X
n=1
λ(n)n−s.
Proposition 3.1. [5] Let L(s, ρ) and N be as above. Let 67 < α < 1. Suppose that L(s, ρ) is zero-free in the rectangle [α, 1] × [−(log N )2, (log N )2]. If N
is sufficiently large, then for any 0 < k < 1−α16 ,
log L(1, ρ) = X p6(log N )k
This implies immediately that L(1, ρ) (log log N )lsince |λ(p)| ≤ l and
P
p≤x1p ∼ log log x.
Due to lack of GRH, we cannot use the above result directly. We extend the result of Kowalski-Michel to isobaric automorphic representations of
GL(n).
Let n = n1+ · · · + nr, and let S(q) be a set of isobaric representations
π = π1 π2· · · πr, where πj is a cuspidal automorphic representation of
GL(nj)/Q and satisfies the Ramanujan-Petersson conjecture at the finite places. We assume that for π, π0 ∈ S(q), πj πj0 for each j. Moreover, S(q) holds the following conditions:
(1) There exists e > 0 such that for π = π1 π2· · · πr ∈ S(q),
Cond(π1) · · · Cond(πr) ≤ qe,
(2) There exists d > 0 such that |S(q)| ≤ qd. (3) The Γ factors of πj are of the formQnj
k=1Γ(s2+ αk), where αk∈ R.
Let, for α ≥ 34, T ≥ 2,
N (π; α, T ) = |{ρ : L(ρ, π) = 0, Re(ρ) ≥ α, |Im(ρ)| ≤ T }|,
(zeros counted with multiplicity). Then clearly, N (π; α, T ) = N (π1; α, T ) +
· · · + N (πr; α, T ).
Theorem 3.1. For some B ≥ 0, X
π∈S(q)
N (π; α, T ) TBqc02α−11−α .
One can choose any c0 > c00, where c00= 5n20e + d and n0 = max{ni}1≤i≤r. Proof. Let S(q)j be the set of the cuspidal automorphic representations consisting of the j-th component of π. Since πj πj0 for each j, |S(q)j| = |S(q)| for all j = 1, 2, · · · , r. Then clearly, Cond(πj) ≤ qe and |S(q)
j| ≤ qd. So X π∈S(q) N (π; α, T ) = X π∈S(q) r X j=1 N (πj; α, T ) = r X j=1 X πj∈S(q)j N (πj; α, T ).
Now we apply the result of Kowalski-Michel [14] to the inner sum. They assumed that the Gamma factors of πj are the same. However, the assump-tion is used only to obtain the convexity bound (Lemma 10 of [14]), and our Γ-factors provide the same convexity bound. Hence our result follows. In the following, we apply the above result to a family of Artin L-functions. In this case, the Γ-factors are a product of Γ(s2) and Γ(s+12 ).
4. D5 extension with signature (1, 2) We use a polynomial in [22] for the case of D5 extension;
f (x, t) = x5− tx4+ (2t − 1)x3− (t − 2)x2− 2x + 1.
Its discriminant is (4t3− 28t2+ 24t − 47)2 and the signature is (1, 2) for
t ≤ 6. (There is a minor mistake in [22]. The discriminant of f (x, t) is given
by 16(4t3− 28t2+ 24t − 47)2 in [22]. But it does not affect his result.) Let
θt be a root of f (x, t). Schöpp found the fundamental units in the equation order Z[θt]. More precisely, he shows
Theorem 4.1 (Schöpp). The elements θt, θt− 1 form a system of
indepen-dent units in the order Z[θt]. Moreover, they are fundamental units in Z[θt]
for t ≤ 6.
However, Schöpp could not show that Z[θt] is the maximal order of Q(θt). Lavallee, Spearman, Williams and Yang [15] found a parametric family of quintics with a power integral basis. The parametric polynomial Fb(x) is given by
Fb(x) := x5− 2x4+ (b + 2)x3− (2b + 1)x2+ bx + 1, b ∈ Z. and its discriminant is (4b3+ 28b2+ 24b + 47)2.
They showed that when 4b3+ 28b2+ 24b + 47 is square-free, then the field Q(θb) generated by a root θb of Fb(x) has a power integral basis Z[θb]. Since x5F−t
1
x
= f (x, t), this implies that θt, θt− 1 are fundamental units of Q(θt) when 4t3− 28t2+ 24t − 47 is square-free.
Schöpp found the locations of roots of f (x, t).
Lemma 4.1 (Schöpp). Let θt(1) be the real root and let θ(2)t =θt(3), θt(4) =
θt(5) be the pairs of complex roots of f (x, t). Then we have the following approximations: (i) −t + 2 +2 t < |θ (1) t | < −t + 2 +1t f or t < −4 (ii) −t + 3 +2 t < |θ (1) t − 1| < −t + 3 + 1t f or t < −4 (iii) 1 2√−t < |θ (2) t | < √2−t f or t < −4 (iv) q1 −4t3 < |θt(2)− 1| <q1 −5t6 f or t < −144 (v) q 1 +3t < |θt(4)| <q1 −t12 f or t < −174 (v) q−5 6t < |θ (4) t − 1| < q −14 13t f or t < −139.
Let Kt be the quintic field by adjoining θ(1)t to Q. Since we know the absolute value of roots, it is easy to show that the regulator RKt of a quintic field Ktis
RKt (log dKt)
when 4t3− 28t2+ 24t − 47 is square-free.
We claim that f (x, t) gives rise to a regular D5 extension over Q(t),
i.e., if we consider f (x, t) as a polynomial over Q(t) and E is the splitting field, then E ∩ Q = Q. This is equivalent to the fact that Gal(EQ/Q(t)) '
Gal(E/Q(t)).
By [13], page 41, the Weber sextic resolvent of f (x, t) is
G(z) = (z3+ b4z2+ b2z + b0)2− 210(4t3− 28t2+ 24t − 47)2z, where b4 = −4t2 − 4t + 37, b2 = 64t3− 312t2 + 328t + 115, b0 = 64t3 − 2884t2+ 4348t − 249. It factors as G(z) = (z − 4t2+ 12t − 9) × (z5− (4t2+ 20t − 83)z4 + (128t3− 368t2− 816t + 2346)z3 − (1024t4− 7040t3+ 16536t2+ 3448t − 29126)z2 + (2048t4+ 51072t3− 201328t2+ 18640t + 256933)z − (1024t4− 89216t3+ 1948548t2− 231404t + 6889)).
Therefore, the Galois group of f (x, t) over Q(t) and Q(t) is either D5 or
C5. In order to distinguish it, we use the criterion in [13], page 42. Namely,
the Galois group is C5 if and only if the resolvent R(x1− x2, f (x, t))(X)
factors into irreducible polynomials of degree 5. Here
R(x1− x2, f (x, t))(X) = X−5Res(f (Y − X, t), f (Y, t)) = (X10+ X8(−2t2+ 12t − 8) + X6(t4− 12t3+ 46t2− 56t − 4) + X4(−2t4+ 16t3− 8t2− 112t + 127) + X2(t4− 60t2+ 128t + 6) − 4t3+ 28t2− 24t + 47) × (X10+ X8(−2t2+ 8t − 2) + X6(t4− 8t3+ 20t2− 4t + 7) + X4(−2t4+ 36t2− 4t + 41) + X2(t4− 4t3+ 66t2− 36t + 103) − 4t3+ 28t2− 24t + 47)
It is clear that the above factors cannot be factored into irreducible polynomials of degree 5. Hence the Galois group of f (x, t) over both Q(t) and Q(t) is D5.
From now on, we replace t by −t in f (x, t) because we want that t varies in positive integers. Then the discriminant of f (x, t) become (4t3+ 28t2+
24t + 47)2. Let ˆK
t be the Galois closure of Kt and Gal( ˆKt/Q) ' D5. Then
ζKt(s)
where ρt= σt,1⊕σt,2and σ1,t and σ2,tare the 2-dimensional representations
of D5. Now D5 has the cyclic subgroup C5 of order 5 and let Mt be the fixed field by C5. Then σ1,t, σ2,t are induced by non-trivial characters for
c
Kt/Mt. Hence the Artin conductors of σ1 and σ2 equal |dMt|NMt/Q(bt)
where bt is the Artin conductor of nontrivial characters of C5. Since dKt = (4t3+ 28t2+ 24t + 47)2 = (|d
Mt|NMt/Q(bt))
2 for 4t3+ 28t2+ 24t + 47
square-free, the Artin conductors of σ1,t, σ2,t are both 4t3+ 28t2+ 24t + 47. Hence,
the Hypotheses of Theorem 3.1 are satisfied.
By Serre’s observation (see p.45 in [23]), there exists a constant cf de-pending only on f such that for every prime q ≥ cf, there is an integer tq with q splits completely in Ktfor all t ≡ tq mod q.
Now for given X 1, define y = log log Xlog X and M =Q
cf≤q≤yq. Let tM be
an integer such that tM ≡ tq modulo q for all cf ≤ q ≤ y. Here log M ∼ y, and hence M X for any > 0.
Since discriminant discf (x, t) of f (x, t) is a polynomial of degree 6, there is a constant C > 0 such that discf (x, t) < Ct6. Now we define a set L(X)
of square-free integers,
L(X) = {X
2 < t < X | 4t
3+ 28t2+ 24t + 47 square-free, t ≡ t
Mmod M }. Let t = M m + tM, and let h(m) = 4(M m + tM)3 + 28(M m + tM)2 + 24(M m + tM) + 47. It is a cubic polynomial. By [10], page 69, |L(X)| =
β2MX + O( X M (logMX)12
) for some constant β. Hence |L(X)| X1−. By the construction of L(X), every t ∈ L(X) gives rise to a distinct automorphic
L-function L(s, ρt) = P∞n=1λt(n)n−s of GL(4)/Q with λt(q) = 4 for all
cf ≤ q ≤ y.
By the above argument, we have
X1− | L(X) | X.
We apply Theorem 3.1 to the family L(X) with n0= 2, T = (log CX6)2, e = 6 and d = 1. Let c0 = 31. Choose α with c02α−11−α < 10098. Then every
L(s, ρt) in L(X) excluding exceptional O(X98/100) L-functions is zero free in the rectangle [α, 1] × [−(log dKt)
2, (log d Kt) 2]. Applying Proposition 3.1 to L(s, ρt), we have log L(1, ρt) = X q6(log dKt)1/2 λ(q)q−1+ O(1) = X cf6q6(log dKt)1/2
4q−1+ O(1) = 4 log log log dKt+ O(1),
where we used the fact that (log dKt)
1/26 y = log X
log log X for large X. So we
have L(1, ρt) (log log dKt)
regulator RKt, we have the required result hKt dKt
1/2(log log dKt)
4
(log dKt)2 .
We summarize our result as follows:
Theorem 4.2. There is a constant c > 0 such that there exist K ∈
K(5, D5, 1, 2) with arbitrarily large discriminant dK for which
hK > cd1/2K
(log log dK)4 (log dK)2 .
Remark 4.1. When t > 7, f (x, t) = x5−tx4+(2t−1)x3−(t−2)x2−2x+1
gives rise to a totally real extension Ktwhose Galois closure has D5 as the
Galois group. In this case, numerical calculation shows that the regulator of Ktis quite large: If t = 103, R Kt ∼ 8 × 10 7; if t = 104, R Kt ∼ 2.3 × 10 10; if t = 105, RKt ∼ 9.3 × 10 13. 5. D4 extension
Nakamula [19] constructed quartic fields with small regulators whose Galois closures have D4 as the Galois group. We prove that Nakamula’s family of quartic fields have large class numbers. We conjecture that they have the largest possible class numbers. (See Remark 7.3.) Nakamula uses a polynomial with 3 parameters
f = x4− sx3+ (t + 2u)x2− usx + 1 where (s, t, u) ∈ N × Z × {±1}, (s, t, u) 6= (1, −1, 1).
The discriminant Df of f is given by
Df = D12D2 with D1= s2− 4t, D2= (t + 4u)2− 4us2.
For a zero of f with || ≥ 1, we define α := + u−1. Put
K = Q(), F = Q(pD1), L = Q( p D2), M = Q( p D1D2) Then F = Q(α) ⊆ K = F () = F (√α2− 4u).
With signs of D1 and D2 we can determine the signature of K. More
precisely,
Lemma 5.1. [19] Assume F 6= Q and L 6= Q. Then K is a non-CM quartic field with a quadratic subfield F , and || > 1. If F = L, then K is cyclic over Q. If F 6= L, then K is non-Galois over Q, and the composite M K is dihedral over Q and cyclic over M . Moreover
(r1, r2) = (0, 2) if D1 < 0 (r1, r2) = (2, 1) if D2 < 0 (r1, r2) = (4, 0) otherwise.
Note that if K is not totally complex, the quadratic subfield F is real. Let K = M K be the Galois closure of K and G = Gal(c K/Q) is iso-c
morphic to D4. Then G has a subgroup H isomorphic to C2 such that
c
KH = K. Let IndGH1H = 1 + ρ be the induced representation of G by the trivial representation of H where ρ is a 3-dimensional representation of D4.
Here ρ is no longer irreducible but a sum of the non-trivial 1-dimensional representation χ and the 2-dimensional representation ψ of D4. Since ψ is
modular, ρ is modular. We can check easily that the Artin conductor of χ is the absolute value of discriminant dF of the quadratic subfield F of K and the Artin conductor of ρ equals to
dK dF . Then L(s, ρ) = L(s, χ)L(s, ψ) = ζK(s) ζ(s) .
5.1. D4 extension with the signature (0,2). We specify that s = u = 1. Then we have, for positive integer t,
f (x, t) = x4− x3+ (t + 2)x2− x + 1
with D1 = 1−4t, D2= t2+8t+12. If D1, D2are square-free for odd integer
t, DK equals (1 − 4t)2(t + 2)(t + 6). For a positive integer t, D1 is negative,
by Lemma 5.1, (r1, r2) = (0, 2) and M K/Q is a D4 Galois extension.
Nakamula estimated the regulator RK of the field K.
RK = 1 4log
dK
16 + o(1) as dK → ∞.
To show that f (x, t) gives rise to a regular D4extension, we briefly recall
how to determine the Galois group of a quartic polynomial over an arbitrary field F in [3], page 358. We write a quartic polynomial f in the form
f = x4− c1x3+ c2x2− c3x + c4
and we define the Ferrari resolvent of f to be
θf(y) = y3− c2y2+ (c1c3− 4c4)y − c23− c21c4+ 4c2c4.
Theorem 5.1. Let F have characteristic 6= 2, and f ∈ F [x] be monic and irreducible of degree 4. Then Galois group of f over F is determined as follows:
(a) If θf(y) is irreducible over F , then
G =
S4, if disc(f ) /∈ F2
A4, if disc(f ) ∈ F2
(c) If θf(y) has a unique root β in F , then G is isomorphic to D4, if 4β + c21− 4c2 6= 0 and disc(f )(4β + c21− 4c2) /∈ (F∗)2 or 4β + c21− 4c2= 0 and disc(f )(β2− 4c4) /∈ (F∗)2 C4, otherwise.
The Ferrari resolvent of f (x, t) is
y3− (t + 2)y2− 3y + 4t + 6 = (y − 2)(y2− ty − (2t + 3)).
Then disc(f (x, t))(4β + c21− 4c2) = (1 − 4t)3(t + 2)(t + 6) /∈ (Q(t)∗)2 and
(Q(t)∗)2. By Theorem 5.1, the Galois group of f (x, t) over both Q(t) and Q(t) is D4. Hence f (x, t) gives rise to a D4 regular extension over Q(t).
Hence as in the case of D5 extension, we can define M , tM and a set
L(X) of square-free integers L(X) = {X
2 < t < X | (1 − 4t)(t + 2)(t + 6) square-free, t ≡ tM mod M }. For t ∈ L(X), |1 − 4t| is the Artin conductor of the one-dimensional rep-resentation and |(t + 2)(t + 6)(1 − 4t)| is the Artin conductor of the two-dimensional representation. Hence we can apply Theorem 3.1 to L(X).
To estimate |L(X)|, we introduce Nair’s work [18]. For an polynomial
f (x) ∈ Z[x] of degree d, we define, Nk(f, x, h) = Nk(x, h) = |{n : x < n ≤ x + h|f (n) : k-free}|. He showed Theorem 5.2 (Nair). If f (x) = m Y i=1 (aix − bi)αi and α = max i αi, then Nk(x, h) = Y p 1 −ρ(p 2) p2 ! h + O h (log h)k−1 for h = x(α/2k)+ if k > α and > 0. Theorem 5.2 implies |L(X)| = Y p-M 1 −ρ(p 2) p2 ! X 2M + O X M (logMX) ! X1−.
Let c0 = 21. Choose α with c02α−11−α < 10098. By applying Theorem 3.1 to
L(X) with e = 4, d = 1 and T = (log CX4)2, every automorphic L-function excluding exceptional O(X98/100) L-functions has a zero-free region [α, 1] ×
[−(log |dKt|)
2, (log |d
Kt|)
2]. Applying Proposition 3.1 to L(s, ρ
t) having the desired zero-free region, we have
log L(1, ρt) = X q6(log dKt)1/2 λ(q)q−1+ O(1) = X cf6q6(log dKt)1/2
3q−1+ O(1) = 3 log log log dKt+ O(1),
where we used the fact that (log dKt)
1/26 y = log X
log log X for large X. So we
have L(1, ρt) (log log dKt)
3.
By the class number formula and the size of regulator RKt, we have the required result hKt dKt 1/2(log log dKt) 3 (log dKt) . We summarize as follows:
Theorem 5.3. There is a constant c > 0 such that there exist K ∈
K(4, D4, 0, 2) with arbitrarily large discriminant dK for which
hK > cd1/2K
(log log dK)3 (log dK)
.
5.2. D4 extension with the signature (2, 1). We specify that u = 1, t = 1. Then we have f (x, s) = x4− sx3+ 3x2− sx + 1 and D
1 = s2− 4
and D2= 25 − 4s2= (5 + 2s)(5 − 2s). Assume that D
1 and D2 are
square-free for odd integers s. Then dF = D1, dL= D2 and by Lemma 5.1 we have
dK = Df = (s2− 4)2(5 + 2s)(5 − 2s). For a positive integer s bigger than 3, D1 is positive and D2 is negative, by Lemma 5.1, (r1, r2) = (2, 1) and
M K/Q is a D4 Galois extension.
Nakamula showed for the field generated by f (x, t),
QRK RF = 1 3log |dK| 4 + o(1), RF = 1 2log dF + o(1)
as |dK| and dF −→ ∞. Here Q is 1 or 2 depending on K and F . Hence
RK (log |dK|)2.
The Ferrari resolvent of f (x, s) is
y3− 3y2+ (s2− 4)y − 2(s2− 6) = (y − 2)(y2− y + (s2− 6)). Then disc(f (x, s))(4β + c21− 4c2) = (s2 − 4)3(5 + 2s)(5 − 2s) /∈ (Q(s)∗)2
and (Q(s)∗)2. Hence f (x, s) gives rise to a D4 regular extension over Q(s). So we can define M , sM and a set L(X) of square-free integers:
L(X) = {X
2 < s < X | (s
2− 4)(25 − 4s2) square-free, s ≡ s
For s ∈ L(X), s2−4 is the Artin conductor of the one-dimensional represen-tation and |(s2−4)(25−4s2)| is the Artin conductor of the two-dimensional
representation. Hence we can apply Theorem 3.1 to L(X). Theorem 5.2 implies |L(X)| = Y p-M 1 −ρ(p 2) p2 ! X 2M + O X M (logMX) ! X1−, and we have
Theorem 5.4. There is a constant c > 0 such that there exist K ∈
K(4, D4, 2, 1) with arbitrarily large discriminant dK for which
hK > cd1/2K
(log log |dK|)3
(log |dK|)2
.
5.3. D4 extension with the signature (4,0). we specify that u = −1, t = 1 and s > 6. Then we have f (x, s) = x4 − sx3 − x2 + sx + 1
and D1 = s2− 4 and D
2 = 9 + 4s2. Assume that D1and D2 are square-free
for odd integer s. Then dF = D1, dL = D2 and by Lemma 5.1 we have
dK = Df = (s2− 4)2(9 + 4s2). For square-free s2− 4 and 4s2+ 9, K, L are always distinct. Hence, by Lemma 5.1, (r1, r2) = (4, 0) and M K/Q is a D4
Galois extension.
Nakamula showed for the field generated by f (x, t),
QRK RF = 1 18log dK 4 log dF 210 + o(1) RF = 1 2log dF + o(1)
as dK and dF −→ ∞. Here Q is 1 or 2 depending on K and F . Hence RK (log |dK|)3. The Ferrari resolvent of f (x, s) is
y3+ y2− (s2+ 4)y − 2(s2+ 2) = (y + 2)(y2− y − (s2+ 2)).
Then disc(f (x, s))(4β + c21 − 4c2) = (s2 − 4)3(9 + 4s2) /∈ (Q(s)∗)2 and
(Q(s)∗)2. Hence f (x, s) gives rise to a D4 regular extension over Q(s). We can define M , sM a set L(X) of square-free integers:
L(X) = {X
2 < s < X | (s
2− 4)(4s2+ 9) square-free, s ≡ s
M mod M }. Since 4s2+ 9 is irreducible, we cannot apply Theorem 5.2 to L(X). Nair also showed
Theorem 5.5 (Nair). If f (x) = Qm
i=1(fi(x))αi ∈ Z[x], where each fi is
irreducible, α = maxiαi and deg fi(x) = gi, then
Nk(x, h) = Y p 1 −ρ(p k) pk ! h + O h (log h)k−1
for h = xθ where 0 < θ < 1 and k ≥ maxi{λgiαi}, (λ = √
2 − 1/2) provided
that at least one gi≥ 2.
Theorem 5.5 implies that |L(X)| X1−. For s ∈ L(X), s2− 4 is the Artin conductor of the one-dimensional representation and (s2− 4)(4s2+ 9)
is the Artin conductor of the two-dimensional representation. Hence we can apply Theorem 3.1 to L(X). We have
Theorem 5.6. There is a constant c > 0 such that there exist K ∈
K(4, D4, 4, 0) with arbitrarily large discriminant dK for which
hK > cd
1/2
K
(log log dK)3 (log dK)3 .
6. D3 extensions with signature (1, 1)
We consider a family of non-abelian cubic fields with signature (1, 1) with the largest possible class numbers. This is different from the one considered in [5]. Our family is the one considered by Ishida [11].
Theorem 6.1 (Ishida). Let Kt= Q(η) be the cubic field of signature (1,1),
where η is the real root of the cubic equation
x3+ tx − 1 = 0, (t ∈ Z, t ≥ 2).
If 4t3+ 27 is square-free or t = 3m and 4m3+ 1 is square-free, then η is
the fundamental unit of K.
It is easy to show that the real root η is located between −1+t and −1t for any > 0. Hence for t with square-free 4t3+ 27, the regulator RKt is
log t < RKt < (1 + ) log t.
Since dKt = −(4t3+ 27), RKt log |dKt|.
Also it is clear that x3+ tx − 1 gives rise to a regular D
3 (= S3) extension
over Q(t). Hence we can define tM and M similarly as before. Define
L(X) = {X
2 < t < X | 4t
3+ 27 square-free and t ≡ t
M mod M } By the work of Hooley as in section 4, we have X1− |L(X)| X. Hence we have
Theorem 6.2. There is a constant c > 0 such that there exist K ∈
K(3, D3, 1, 1) with arbitrarily large discriminant dK for which
hK > cd1/2K
(log log dK)2 (log dK)
7. Cyclic extensions
Let f (x, t) = xn+ a1(t)xn−1+ · · · + an−1(t)x ± 1 be an irreducible poly-nomial over Q(t) such that ai(t) ∈ Z[t]. Suppose f (x, t) gives rise to a cyclic extension over Q(t), and if t ∈ Z, it gives rise to a totally real ex-tension over Q. For each integer t > 0, let Kt be the cyclic extension over Q. Let Gal(Kt/Q) = {1, σ, σ2, ..., σn−1}. Let θ be a root of f (x, t). Then
θ, σ(θ), ..., σn−1(θ) are roots of f (x, t).
We show that if n = p is a prime, σ(θ), ..., σp−1(θ) form independent units, and the regulator of Kt is small. By definition, the regulator of Z[θ] is
R = | det(log |σi+j(θ)|)1≤i,j≤p−1|.
Theorem 7.1. (1) R 6= 0, and (2) R (log t)p−1. Proof. By Lemma 5.26 in [26], we have
R = 1 p Y χ6=1 p−1 X i=0 χ(σi) log |σi(θ)| ,
where the product runs over the nontrivial characters of Gal(Kt/Q). Since t−c σi(θ) td for some c, d > 0 depending only on f (x, t), | log |σi(θ)|| log t. Hence (2) follows.
Since θ · σ(θ) · · · σp−1(θ) = ±1,
log |θ| + log |σ(θ)| + · · · + log |σp−1(θ)| = 0. Hence we write p−1 X i=0 χ(σi) log |σi(θ)| = p−1 X i=1 (χ(σi) − 1) log |σi(θ)|.
Since p is a prime, by [7], σ(θ), . . . , σp−1(θ) are multiplicatively indepen-dent. Hence log |σ(θ)|, . . . , log |σp−1(θ)| are linearly independent over Q. By Baker’s theorem [1], they are linearly independent over Q.
Since χ(σi) are roots of unity and χ 6= 1, one of χ(σi) − 1 is not zero. Hence p−1 X i=1 (χ(σi) − 1) log |σi(θ)| 6= 0.
Remark 7.1. If n is not a prime, we still have R (log t)n−1. However,
R = 0 for a composite number n. We show this for simplest quartic and
sextic fields. Duke [4] proved that cyclic cubic fields given by f (x, t) =
x3− tx2− (t + 3)x − 1 (studied by D. Shanks) have the smallest possible
7.1. Cyclic quartic fields. Consider totally real cyclic quartic fields Kt generated by a root of
f (x, t) = x4− tx3− 6x2+ tx + 1, t ∈ Z+.
Here Disc(f (x, t)) = 4(t2+ 16)3.
We can express the 4 roots of f (x, t) explicitly.
θ1,2,3,4= ± 4 √ t2+ 16 q√ t2+ 16 ± t 2√2 ± √ t2+ 16 4 + t 4, t 6= 0, 3 where the second and third ambiguous signs agree.
Let θ1 be the largest root by choosing + for all signs. The Galois group
action on the roots is given by
σ : θj 7→
θj− 1
θj+ 1
= θj+1, j = 1, 2, 3, 4.
Now θ1, θ2= σ(θ1) and θ3 = σ2(θ1) are not multiplicatively independent.
The regulator R = 1 4 Y χ6=1 3 X i=0 χ(σi) log |σi(θ)| !
vanishes because the term corresponding to χ(σ) = eπi is zero.
It is known that θ1, θ2 and tare independent units where t is the fun-damental units of Q(√t2+ 16) (See p.10 in [16]). When t is even, we can
find t.
Proposition 7.1 (Lazarus). When t is even, t is given by
t= t/2+√(t/2)2+4 2 , t ≡ 2 mod 4 1+√5 2 , t = 8 (t/4) +p (t/4)2+ 1, otherwise.
We replace t in f (x, t) by 2t because it is convenient to consider only even t. Then
f (x, t) = x4− 2tx3− 6x2+ 2tx + 1
with disc(f (x, t)) = 28(t2+ 4)3. When t2+ 4 is square-free, the field dis-criminant dKt equals 24(t2+ 4)3, and Kt has the unique quadratic subfield
Mt = Q( √
t2+ 4). Let H ' C
2 be the unique subgroup of order 2 in C4.
Then IndC4
H 1H = 1 + χ2 where χ the generator of the group of charac-ters for C4. Hence the Artin conductor f (χ2) of χ2 equals t2 + 4 when
t2+ 4 is square-free. Since dKt = f (χ)f (χ2)f (χ3) and χ3 = χ, we have f (χ) = f (χ3) = 22(t2+ 4).
Hence by Proposition 7.1, when t2+ 4 is square-free, the regulator RKt is given by RKt log 3t. Since d Kt = 2 4(t2+ 4)3, RKt (log dKt) 3.
From the root formula for f (x, t), it is clear that f (x, t) gives rise to a C4
regular extension over Q(t). Let
L(X) = {X
2 < t < X | t
2+ 4 square-free, t ≡ t
Mmod M }
where tM and M are defined similarly as before. Then we have X1−
|L(X)| X. Since the Artin conductors of the characters for the simplest quartic fields are increasing functions in t, they satisfy the hypothesis in Theorem 3.1. Hence we have
Theorem 7.2. There is a constant c > 0 such that there exist K ∈
K(4, C4, 4, 0) with arbitrarily large discriminant dK for which
hK > cd 1 2 K (log log dK)3 (log dK)3 .
7.2. Cyclic quintic fields. Emma Lehmer [17] introduced a family of
quintic polynomials f (x, t) for t ∈ Z:
f (x, t) = x5+ t2x4− (2t3+ 6t2+ 10t + 10)x3
+ (t4+ 5t3+ 11t2+ 15t + 5)x2+ (t3+ 4t2+ 10t + 10)x + 1. It is easy to show that f (x, t) is irreducible for all t ∈ Z when we observe it modulo 2. And it is also known that the zeros of f (x, t) generate a cyclic extension Kt of degree 5 over Q. Here Disc(f (x, t)) = (t3 + 5t2+ 10t + 7)2(t4+ 5t3+ 15t2+ 25t + 25)4.
Let G = Gal(Kt/Q) be the Galois group and σ be a generator of G given by
σ(θ) = (t + 2) + tθ − θ
2
1 + (t + 2)θ
for a root θ of f (x, t). For the details, we refer to [21]. Also it is obvious that the Galois groups of f (x, t) over Q(t) and over Q(t) are both C5 generated
by σ. Hence f (x, t) gives rise to a regular C5 extension over Q(t).
Schoof and Washington studied these simplest quintic fields when Pt =
t4+ 5t3+ 15t2 + 25t + 25 is a prime number. When Pt is a prime, then the zeros of f (x, t) form a fundamental system of units in Kt. Gaál and Pohst extended this result for square-free Pt(see the proof of Theorem 3.5 in [21]).
In this case, by Theorem 7.1, RKt (log t)4. It also follows from [21]: Let
U denote the group of units generated by the zeros modulo {±1}. Define iθ = [O∗Kt/{±1} : U ]. Schoof and Washington [21] showed that iθ ≤ 11 if
|t + 1| ≥ 20 and
R = |det(log |σi+j(θ)|)1≤i,j≤4| ≤
71 + 36
log |t + 1|
Jeannin [12] found the prime factorization of Pt.
Theorem 7.3 (Jeannin). The number Pt is written in a unique way: Pt = 5cq5Qni=1p
xi
i , c ∈ {0, 2}, q ∈ N, pi distinct primes, xi ∈ [1, 4]. So
the conductor of Kt is ft= 5cQni=1pi.
Especially if Pt is cube-free, then Pt= 5cQni=1p xi i and x1 = 1 or 2 and t4 Pt≤ 5c( n Y i=1 pi)2 ≤ dKt
since dKt = ft4. Hence for cube-free Pt, when we combine all these argu-ments, we have
RKt log
4(d
Kt).
Let L(X) be a finite set given by
L(X) = {X
2 < t < X | Pt cube-free and t ≡ tM modM }. Then by [10], page 69, we have X1− |L(X)| X.
For cyclic extensions of prime degree, the conductor of a cyclic extension equals the Artin conductors of characters for the extension. We showed that the product of prime divisors of Pt= t4+ 5t3+ 15t2+ 25t + 25 is the conductor. But it would be possible that the sets of distinct prime divisors of
Ptcoincide for different t’s. Let ν(n) is the number of distinct prime divisors of n. For each t, the number of possible repetition is bounded by 2ν(Pt)
because we assume that Ptis cube-free. It is known that ν(n) log log nlog n .(See page 167 in [20]) Hence, for all t < X, 2ν(Pt) 2
log X
log log X X. After
removing the possible repetition, we can say that the Artin conductors are distinct.
Hence we have
Theorem 7.4. There is a constant c > 0 such that there exist K ∈
K(5, C5, 5, 0) with arbitrarily large discriminant dK for which
hK > cd
1/2
K
(log log dK)4 (log dK)4 .
7.3. Cyclic sextic fields. It was Gras [9] who introduced the simplest
sextic polynomial f (x, t) first, given by
f (x, t) = x6− t − 6 2 x 5− 5t + 6 4 x 4− 20x3+ 5t − 6 4 x 2+t + 6 2 x + 1 and discriminant of f (x, t) is 23146 (t2+ 108)5.
Let Kt = Q(θ), where θ is a root of f (x, t). She showed the following properties:
(1) If t ∈ Z − {0, ±6, ±26}, then f (x, t) is irreducible in Q[X], and
Kt is a real cyclic sextic field; a generator σ of its Galois group is characterized by the relation σ(θ) = (θ − 1)/(θ + 2). We have
K−t = Kt for all t ∈ Z, and we can suppose that t ∈ N − {0, 6, 26}. (2) The quadratic subfield of Kt is k2 = Q(√t2+ 108).
(3) The cubic field of Ktis k3 = Q(φ), where
φ = θ−1−σ3 = − 2θ + 1 θ(θ + 2) and Irr(φ, Q) = x3−t − 6 4 x 2−t + 6 4 x − 1;
the discriminant of this polynomial is ((t2+108)/16)2. If t ≡ 2 (mod 4), k3 is the simplest cubic field.
(4) The conductor f of Kt is given by the following procedure: Let m be the product of primes, different from 2 and 3, dividing t2+ 108
with an exponent not congruent to 0 modulo 6. Then f = 4k3lm,
where k = ( 0, t ≡ 1 (mod 2) or t ≡ ±6 (mod 16) 1, otherwise , l = 0, t ≡ 1 (mod 3) 1, t ≡ 0 (mod 27) 2, otherwise As in the case of simplest quartic fields, θ, σ(θ), σ2(θ), σ3(θ) and σ4(θ) do not form independent units. The regulator
R = 1 6 Y χ6=1 5 X i=0 χ(σi) log |σi(θ)| !
vanishes because the term corresponding to χ(σ) = eπi is zero. Let
S(X) =n0 < r < X | (3r2+ 3r + 1)(12r2+ 12r + 7) square-freeo.
For all r ∈ S(X), let t = (6r + 3)(36r2+ 36r + 18) and we consider fields
Lr = Kt = Q(w) where w = θ1−σ
3
= −θ(2θ+1)θ+2 . We note that t2+ 108 = 432(12r2 + 12r + 7)(3r2 + 3r + 1)2. Then there exits a unit v such that
w = v1+σ. Hence v = (w+1)− √
(w+1)2−8w
2 . Gras showed that if r ∈ S(X),
then the conductor of k2 is f2= 36r2+ 36r + 21, and the fundamental unit of k2 is
2 =
(12r2+ 12r + 5) + (2r + 1)√36r2+ 36r + 21
2 .
By Proposition 1 in [27], the conductor of k3 is (3r2 + 3r + 1)(12r2 +
12r + 7). By the above property (4), if r ∈ S(X), the conductor f6 of Ktis 3(12r2+ 12r + 7)(3r2+ 3r + 1).
Let σ be the generator of Gal(Kt/Q) ' C6, k3 = K<σ
3>
t and k2 =
Kt<σ2>. Let χ be the generator of the group of characters for Gal(Kt/Q) with χ(σ) = e2πi/6. Then Ind<σ>
<σ2>1<σ2> = 1<σ>+ χ3, Ind<σ><σ3>1<σ3> =
1<σ>+ χ2+ χ4 and Ind<σ><σ3>ϕ = 1<σ>+ χ + χ5 where ϕ is the non-trivial character for < σ3 >. Hence the Artin conductor of χ3 equals the field discriminant of k2, which is 3(12r2+ 12r + 7), and the Artin condutors of
χ2 and χ4 are both (3r2+ 3r + 1)(12r2+ 12r + 7). The Artin conductors of
χ and χ5 equals (3r2+ 3r + 1)(12r2+ 12r + 7)pN (b) where b is the Artin
conductor of ϕ. Since the Artin conductors of χ and χ5 are divisors of f6,
p
N (b) is at most 3. Hence we verified that the hypothesis of Theorem 3.1
is satisfied.
Since t = (6r + 3)(36r2+ 36r + 18) ≡ 2 mod 4, the field k3 is a simplest
cubic field. Hence for r ∈ S, we have an explicit system of fundamental units: {2, τ, τσ, v, vσ} where τ is a root of x3−t−6 4 x 2−t+6 4 x − 1. Hence for t = (6r + 3)(36r 2+
36r + 18) with r ∈ S(X), the regulator RKt (log dKt)
5 .
Now we show that f (x, (6r + 3)(36r2+ 36r + 18)) gives rise to a regular
C6 extension over Q(r). If θr is a root of f (x, (6r + 3)(36r2+ 36r + 18)),
then it is clear Q(r)(θr) is the splitting field of f (x, (6r + 3)(36r2+ 36r + 18) over Q(r) with Galois group C6=< σ >. By the same argument, the Galois
group of f (x, (6r + 3)(36r2+ 36r + 18)) over Q(t) is also C6 =< σ >. Hence the claim follows.
Now define L(X) = X 2 < r < X (3r2+ 3r + 1)(12r2+ 12r + 7) square-free, r ≡ rM mod M
where rM and M are defined similarly as before and we can show that
X1− |L(X)| X. Hence we have
Theorem 7.5. There is a constant c > 0 such that there exist K ∈
K(6, C6, 6, 0) with arbitraril large discriminant dK for which
hK > cd1/2K
(log log dK)5 (log dK)5
.
Remark 7.2. We could not find in the literature a family of polynomials
of degree 7, f (x, t) = x7+ a1(t)x6+ · · · + a6(t)x ± 1, where ai(t) ∈ Z[t] which
generate cyclic extensions of degree 7. By Theorem 7.1, such a family would provide a family of cyclic extensions of degree 7 with the largest possible class numbers.
Remark 7.3. Silverman [25] obtained a lower bound of regulator RK of number fields K: RK (log |dK|)r−r0, where r = r
1+ r2− 1 and r0 is the
where f (x, t) gives rise a regular Galois extension over Q(t), the regulator
RKt satisfies
RKt (log |dKt|)
r.
We conjecture that this is always sharp in the case of regular Galois ex-tensions. The CM field Kt = Q(
√
α − t), where α is an algebraic integer,
given in [6], does not give rise to a regular Galois extension over Q(t), since Q(α) ⊂ Kt.
Remark 7.4. Shen [24] considered a parametric polynomial f (x, t)
gener-ating real cyclic octic fields:
f (x, t) = x8− 8tx7− 28x6+ 56tx5+ 70x4− 56tx3− 28x2+ 8tx + 1. However, this polynomial does not give rise to a regular extension over Q(t), since Q(√2) ⊂ Kt, where Kt is the splitting field of f (x, t) (See Theorem 1 in [24]). So this polynomial is not suitable for our purpose.
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Peter J. Cho
Department of Mathematics University of Toronto Toronto ON M5S 2E4 Canada
E-mail: jcho@math.toronto.edu
Henry H. Kim
Department of Mathematics University of Toronto
Toronto ON M5S 2E4 CANADA and Korea Institute for Advanced Study
